If the circle were centered at the origin, of radius r, then r(cos$\theta$, sin$\theta$) traverses the circle once counterclockwise, for 0 $\le$$\theta$$\le$2$\pi$.
What if the circle were centered at, say, (x,y) = (5,2)? Also of radius r.
It's not r($cos\theta$ – 5, $sin\theta$-2) … right?
I'm thinking of the equation for the circle, which would give $(x-5)^2+(y-2)^2=r^2$.
Thanks,
Best Answer
If the circle $\Gamma$ has equation $(x-a)^2+(y-b)^2=R^2$, then $$x=a+R\cos\alpha, y=b+R\sin\alpha$$ parametrize the curve $\Gamma$. Indeed, $$x=a+R\cos\alpha, y=b+R\sin\alpha \Rightarrow \cos\alpha=\frac{x-a}{R} , \sin\alpha=\frac{y-b}{R}$$ but, $$\cos^2\alpha+\sin^2\alpha=1 \Rightarrow \frac{(x-a)^2}{R^2}+\frac{(y-b)^2}{R^2}=1 \Rightarrow (x-a)^2+(y-b)^2=R^2 $$